$\dfrac{d}{dx}\left(\dfrac{\sin(x)}{\cos(x)}\right)=$
Answer: $\dfrac{\sin(x)}{\cos(x)}$ is the quotient of two, more basic, expressions: $\sin(x)$ and $\cos(x)$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{\sin(x)}{\cos(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sin(x))\cos(x)-\sin(x)\dfrac{d}{dx}(\cos(x))}{(\cos(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\cos(x)\cdot \cos(x)-\sin(x)\cdot (-\sin(x))}{\cos^2(x)}&&\gray{\text{Differentiate }\sin(x)\text{ and }\cos(x)} \\\\ &=\dfrac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}&&\gray{\text{Simplify}} \\\\ &=\dfrac{1}{\cos^2(x)}&&\gray{\text{Pythagorean Identity}} \\\\ &=\sec^2(x)&&\gray{\text{Reciprocal Identity}} \end{aligned}$ Note: Since $\dfrac{\sin(x)}{\cos(x)}=\tan(x)$, we have just shown that $\dfrac{d}{dx}(\tan(x))=\sec^2(x)$. In conclusion, $\dfrac{d}{dx}\left(\dfrac{\sin(x)}{\cos(x)}\right)=\sec^2(x)$ or any other equivalent form.